Probability limits of random variable sums












1












$begingroup$


I have $X_1, X_2, X_3, cdots$ which are independent random variables with the same non-zero mean ($mune0$) and same variance $sigma^2$.



I would like to compute $$lim_{ntoinfty} P[frac{1}nsum^n_{i=1}X_i < frac{mu}{2}]$$ for $mu<0$ and $mu>0$.



My initial thought was to use the central limit theorem but it indicates the variables must be identically distributed which I dont have here -- only have first and second order moments are similar.



Any thoughts on how to start tackling this?






convergence central-limit-theorem probability-limit-theorems







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I have $X_1, X_2, X_3, cdots$ which are independent random variables with the same non-zero mean ($mune0$) and same variance $sigma^2$.



    I would like to compute $$lim_{ntoinfty} P[frac{1}nsum^n_{i=1}X_i < frac{mu}{2}]$$ for $mu<0$ and $mu>0$.



    My initial thought was to use the central limit theorem but it indicates the variables must be identically distributed which I dont have here -- only have first and second order moments are similar.



    Any thoughts on how to start tackling this?






    convergence central-limit-theorem probability-limit-theorems







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I have $X_1, X_2, X_3, cdots$ which are independent random variables with the same non-zero mean ($mune0$) and same variance $sigma^2$.



      I would like to compute $$lim_{ntoinfty} P[frac{1}nsum^n_{i=1}X_i < frac{mu}{2}]$$ for $mu<0$ and $mu>0$.



      My initial thought was to use the central limit theorem but it indicates the variables must be identically distributed which I dont have here -- only have first and second order moments are similar.



      Any thoughts on how to start tackling this?






      convergence central-limit-theorem probability-limit-theorems







      share|cite|improve this question









      $endgroup$




      I have $X_1, X_2, X_3, cdots$ which are independent random variables with the same non-zero mean ($mune0$) and same variance $sigma^2$.



      I would like to compute $$lim_{ntoinfty} P[frac{1}nsum^n_{i=1}X_i < frac{mu}{2}]$$ for $mu<0$ and $mu>0$.



      My initial thought was to use the central limit theorem but it indicates the variables must be identically distributed which I dont have here -- only have first and second order moments are similar.



      Any thoughts on how to start tackling this?






      convergence central-limit-theorem probability-limit-theorems




      convergence central-limit-theorem probability-limit-theorems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 21 '18 at 2:28









      AvedisAvedis

      647




      647






















          2 Answers
          2






          active

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          3












          $begingroup$

          Chebychev's inequality may help.




          For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.







          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            @ClementC. You're right, thanks for catching my mistake.
            $endgroup$
            – angryavian
            Dec 21 '18 at 2:56










          • $begingroup$
            Why is the right side of Chebychev's inquality have n and not n^2?
            $endgroup$
            – Avedis
            Dec 23 '18 at 1:43










          • $begingroup$
            @Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
            $endgroup$
            – angryavian
            Dec 23 '18 at 2:08










          • $begingroup$
            Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
            $endgroup$
            – Avedis
            Dec 23 '18 at 3:51



















          0












          $begingroup$

          Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.



          $E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$



          For $mu<0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (using Chebyshev inequality)



          $$P[|M_n-mu| < -frac{mu}{2}]$$



          $$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.



          For $mu > 0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (flip inequality due to division of both sides by -1)



          $$P[-M_n+mu > frac{mu}{2}]$$



          (using Chebyshev inequality and fact of $|A|=|-A|$)



          $$P[|M_n-mu| > frac{mu}{2}]$$



          $$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
            $endgroup$
            – angryavian
            Dec 23 '18 at 4:20












          • $begingroup$
            You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
            $endgroup$
            – Avedis
            Dec 23 '18 at 14:04













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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Chebychev's inequality may help.




          For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.







          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            @ClementC. You're right, thanks for catching my mistake.
            $endgroup$
            – angryavian
            Dec 21 '18 at 2:56










          • $begingroup$
            Why is the right side of Chebychev's inquality have n and not n^2?
            $endgroup$
            – Avedis
            Dec 23 '18 at 1:43










          • $begingroup$
            @Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
            $endgroup$
            – angryavian
            Dec 23 '18 at 2:08










          • $begingroup$
            Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
            $endgroup$
            – Avedis
            Dec 23 '18 at 3:51
















          3












          $begingroup$

          Chebychev's inequality may help.




          For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.







          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            @ClementC. You're right, thanks for catching my mistake.
            $endgroup$
            – angryavian
            Dec 21 '18 at 2:56










          • $begingroup$
            Why is the right side of Chebychev's inquality have n and not n^2?
            $endgroup$
            – Avedis
            Dec 23 '18 at 1:43










          • $begingroup$
            @Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
            $endgroup$
            – angryavian
            Dec 23 '18 at 2:08










          • $begingroup$
            Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
            $endgroup$
            – Avedis
            Dec 23 '18 at 3:51














          3












          3








          3





          $begingroup$

          Chebychev's inequality may help.




          For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.







          share|cite|improve this answer











          $endgroup$



          Chebychev's inequality may help.




          For $mu > 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| > frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0.$$ For $mu < 0$, we have $$Pleft(frac{1}{n} sum_{i=1}^n X_i ge frac{mu}{2}right) le Pleft(left|frac{1}{n} sum_{i=1}^n X_i - mu right| ge frac{mu}{2}right) le frac{sigma^2/n}{mu^2/4} to 0$$ so $Pleft(frac{1}{n} sum_{i=1}^n X_i < frac{mu}{2}right) to 1$.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 21 '18 at 2:56

























          answered Dec 21 '18 at 2:33









          angryavianangryavian

          41.8k23381




          41.8k23381








          • 1




            $begingroup$
            @ClementC. You're right, thanks for catching my mistake.
            $endgroup$
            – angryavian
            Dec 21 '18 at 2:56










          • $begingroup$
            Why is the right side of Chebychev's inquality have n and not n^2?
            $endgroup$
            – Avedis
            Dec 23 '18 at 1:43










          • $begingroup$
            @Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
            $endgroup$
            – angryavian
            Dec 23 '18 at 2:08










          • $begingroup$
            Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
            $endgroup$
            – Avedis
            Dec 23 '18 at 3:51














          • 1




            $begingroup$
            @ClementC. You're right, thanks for catching my mistake.
            $endgroup$
            – angryavian
            Dec 21 '18 at 2:56










          • $begingroup$
            Why is the right side of Chebychev's inquality have n and not n^2?
            $endgroup$
            – Avedis
            Dec 23 '18 at 1:43










          • $begingroup$
            @Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
            $endgroup$
            – angryavian
            Dec 23 '18 at 2:08










          • $begingroup$
            Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
            $endgroup$
            – Avedis
            Dec 23 '18 at 3:51








          1




          1




          $begingroup$
          @ClementC. You're right, thanks for catching my mistake.
          $endgroup$
          – angryavian
          Dec 21 '18 at 2:56




          $begingroup$
          @ClementC. You're right, thanks for catching my mistake.
          $endgroup$
          – angryavian
          Dec 21 '18 at 2:56












          $begingroup$
          Why is the right side of Chebychev's inquality have n and not n^2?
          $endgroup$
          – Avedis
          Dec 23 '18 at 1:43




          $begingroup$
          Why is the right side of Chebychev's inquality have n and not n^2?
          $endgroup$
          – Avedis
          Dec 23 '18 at 1:43












          $begingroup$
          @Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
          $endgroup$
          – angryavian
          Dec 23 '18 at 2:08




          $begingroup$
          @Avedis Think carefully about what the variance of $frac{1}{n} sum_{i=1}^n X_i$ is.
          $endgroup$
          – angryavian
          Dec 23 '18 at 2:08












          $begingroup$
          Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
          $endgroup$
          – Avedis
          Dec 23 '18 at 3:51




          $begingroup$
          Thank you, I just figure out what you mentioned on my own. However, when I complete the problem, I get a different answer then you do for $mu>0$. Why?
          $endgroup$
          – Avedis
          Dec 23 '18 at 3:51











          0












          $begingroup$

          Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.



          $E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$



          For $mu<0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (using Chebyshev inequality)



          $$P[|M_n-mu| < -frac{mu}{2}]$$



          $$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.



          For $mu > 0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (flip inequality due to division of both sides by -1)



          $$P[-M_n+mu > frac{mu}{2}]$$



          (using Chebyshev inequality and fact of $|A|=|-A|$)



          $$P[|M_n-mu| > frac{mu}{2}]$$



          $$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
            $endgroup$
            – angryavian
            Dec 23 '18 at 4:20












          • $begingroup$
            You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
            $endgroup$
            – Avedis
            Dec 23 '18 at 14:04


















          0












          $begingroup$

          Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.



          $E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$



          For $mu<0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (using Chebyshev inequality)



          $$P[|M_n-mu| < -frac{mu}{2}]$$



          $$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.



          For $mu > 0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (flip inequality due to division of both sides by -1)



          $$P[-M_n+mu > frac{mu}{2}]$$



          (using Chebyshev inequality and fact of $|A|=|-A|$)



          $$P[|M_n-mu| > frac{mu}{2}]$$



          $$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
            $endgroup$
            – angryavian
            Dec 23 '18 at 4:20












          • $begingroup$
            You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
            $endgroup$
            – Avedis
            Dec 23 '18 at 14:04
















          0












          0








          0





          $begingroup$

          Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.



          $E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$



          For $mu<0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (using Chebyshev inequality)



          $$P[|M_n-mu| < -frac{mu}{2}]$$



          $$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.



          For $mu > 0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (flip inequality due to division of both sides by -1)



          $$P[-M_n+mu > frac{mu}{2}]$$



          (using Chebyshev inequality and fact of $|A|=|-A|$)



          $$P[|M_n-mu| > frac{mu}{2}]$$



          $$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.






          share|cite|improve this answer











          $endgroup$



          Define $M_n=frac{1}{n}sum_{i-1}^nX_i$.



          $E[M_n]=mu$, $Var[M_n]=frac{sigma_X^2}{n}$



          For $mu<0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (using Chebyshev inequality)



          $$P[|M_n-mu| < -frac{mu}{2}]$$



          $$P[|M_n-mu| < -frac{mu}{2}] geq 1 - frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity, so the Probability is $1-0=1$.



          For $mu > 0$:



          $$P[M_n<frac{mu}{2}]$$



          $$P[M_n-mu<frac{mu}{2}-mu]$$ (Subtracting $mu$ from both sides)



          $$P[M_n-mu<frac{mu}{2}-mu]$$



          $$P[M_n-mu< -frac{mu}{2}]$$



          (flip inequality due to division of both sides by -1)



          $$P[-M_n+mu > frac{mu}{2}]$$



          (using Chebyshev inequality and fact of $|A|=|-A|$)



          $$P[|M_n-mu| > frac{mu}{2}]$$



          $$P[|M_n-mu| < frac{mu}{2}] leq frac{sigma_X^2/n}{mu^2/4}$$ which goes to zero when n goes to infinity. So, the probability is 0.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 23 '18 at 14:05

























          answered Dec 23 '18 at 3:50









          AvedisAvedis

          647




          647












          • $begingroup$
            In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
            $endgroup$
            – angryavian
            Dec 23 '18 at 4:20












          • $begingroup$
            You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
            $endgroup$
            – Avedis
            Dec 23 '18 at 14:04




















          • $begingroup$
            In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
            $endgroup$
            – angryavian
            Dec 23 '18 at 4:20












          • $begingroup$
            You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
            $endgroup$
            – Avedis
            Dec 23 '18 at 14:04


















          $begingroup$
          In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
          $endgroup$
          – angryavian
          Dec 23 '18 at 4:20






          $begingroup$
          In both cases you should have $(mu/2)^2=mu^2/4$ rather than $mu^2/2$. And it seems our answers agree.
          $endgroup$
          – angryavian
          Dec 23 '18 at 4:20














          $begingroup$
          You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
          $endgroup$
          – Avedis
          Dec 23 '18 at 14:04






          $begingroup$
          You are right and I will fix. I misread your answer. I thought the last statement was a summary of the entire problem, not the last part exclusively. Thank you!
          $endgroup$
          – Avedis
          Dec 23 '18 at 14:04




















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